Because I'm primarily interested in this question from the point of view of $\infty$-categories (in this case, modeled by quasicategories), I'll ask this question using that terminology. In particular, I'll just say "category" instead of $\infty$-category or quasicategory, to keep things legible. I suspect that, up to adding in the word "homotopy," a lot of what I'm going to say holds very generally.

It's not hard to work out that if $Top$ is the category of $\infty$-groupoids (or anima, according to Peter Scholze), then there is an equivalence between comodules over any $X\in Top$, with respect to the Cartesian monoidal structure, and the slice category $Top/X$. I'll write $Comod_X(Top)\simeq Top/X$. Furthermore, by Lurie's straightening/unstraightening (a.k.a. the $\infty$-categorical Grothendieck construction), we have an equivalence $Top/X\simeq Fun(X,Top)$. The first category, $Comod_X(Top)$, is clearly comonadic over $Top$ via the forgetful functor $U\colon Comod_X(Top)→Top$. Moreover, the colimit functor $Fun(X,Top)→Top$ factors through $Comod_X(Top)$ and is the left adjoint of an adjoint equivalence realizing the composite equivalence $Comod_X(Top)?Fun(X,Top)$. Thus, $Fun(X,Top)$ is comonadic over $Top$ via the colimit functor $colim\colon Fun(X,Top)→Top$.

My question is whether or not this is more generally true, perhaps for some generic abstract reasons that I'm not aware or, or not seeing. Note that it is not in general true (I don't think) that $colim:Fun(D,C)→C$ is comonadic for arbitrary categories $D$ and $C$. However, I'm happy with assuming that $D$ is an $∞$-groupoid and that $C$ is "nice," i.e. at least (locally) presentable.

So really the question is, are there check-able conditions under which the colimit functor $colim:Fun(X,C)→C$ is comonadic, under the assumptions that $X$ is an $∞$-groupoid and $C$ is presentable?

  • 3
    $\begingroup$ @AlexanderCampbell I thought this was written somewhere, but it is perhaps just gossip... (oops...) $\endgroup$ – Jonathan Beardsley May 30 at 23:50
  • 3
    $\begingroup$ This seems to fail badly when $C$ is stable. For example if $X=BG$ and $C=Vect$, the functor of coinvariants $Rep(G) \to Vect$ is usually not conservative. I guess you want to restrict to the case where $C$ is a topos or something? $\endgroup$ – Sam Gunningham May 31 at 11:54
  • 2
    $\begingroup$ As is probably clear, your argument also works for $\infty$-toposes and they also have that extra property @DylanWilson mentioned, that $\infty$-groupoid shaped colimite commute with contractible limits. $\endgroup$ – Omar Antolín-Camarena May 31 at 13:39
  • 3
    $\begingroup$ For a concrete example, take $X=BC_2$ and $F$ the functor $BC_2\to Vect$ corresponding to the sign representation of $C_2$ (say over a field of characteristic not 2). Then the colimit of $F$ is just the coinvariants of the sign representation, which vanishes. Or more topologically, the corresponding local system on $BC_2$ has vanishing homology. So the functor is not conservative. One could just as well take the ambient category to be spectra, by embedding $Vect$ inside as $Hk$-modules. $\endgroup$ – Sam Gunningham May 31 at 16:34
  • 2
    $\begingroup$ @TimCampion The situation is not symmetric: $\mathrm{Fun}(X, Top)$ Is equivalent to $Top/X$ via the colimit functor and not via the limit functor. $\endgroup$ – Omar Antolín-Camarena Jun 3 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.